An elastic belt is placed around the rim of a pulley of radius 5 𝑐𝑚. From one point 𝐶 On the belt, the elastic belt is pulled directly away from the centre 𝑂 of the pulley until it is at 𝑃, 10 𝑐𝑚 from the point 𝑂. Find the length of the belt that is still in contact with the pulley. Also, find the shaded area. [use π = 3. 14 and $\sqrt{3}$ = 1. 73]

We know that the tangent of a circle is perpendicular to the radius of the circle through the point of contact.

$\begin{array}{r}\Rightarrow \mathrm{\angle }OAP={90}^{\circ }\\ \Rightarrow \cos \theta =\frac{OA}{OP}\\ \Rightarrow \cos \theta =\frac{5}{10}\end{array}$

$\begin{array}{r}\Rightarrow \cos \theta =\frac{1}{2}\\ \Rightarrow \theta ={\cos }^{-1}\left(\frac{1}{2}\right)\\ \Rightarrow \theta ={60}^{\circ }\\ \begin{array}{c}\mathrm{\angle }AOB={60}^{\circ }+{60}^{\circ }\\ ={120}^{\circ }\end{array}\\ \begin{array}{r}\mathrm{reflex} \mathrm{\angle }AOB=360-\mathrm{\angle }AOB\\ ={360}^{\circ }-{120}^{\circ }\\ ={240}^{\circ }\\ We know that, length of belt in contact with pulley = ADB\\ Length of major arc = \begin{array}{r} \\ \frac{{240}^{\circ }}{{360}^{\circ }}\times 2\times 3.14\times 5\\ =20.93\mathrm{cm}\end{array}\end{array}\\ \end{array}$

Now, in right angled triangle OAP we have:

$\begin{array}{r}\sin {60}^{\circ }=\frac{AP}{OP}\\ \frac{\sqrt{3}}{2}=\frac{AP}{10}\\ AP=5\sqrt{3}\mathrm{cm}\end{array}$

$\begin{array}{r}\text{ Area of triangle }=\frac{1}{2}\times \text{ Base }\times \text{ Height }\\ =\frac{1}{2}\times AP\times OA\\ =\frac{1}{2}\times 5\sqrt{3}\times 5\\ =\frac{25}{2}\times \sqrt{3}{\mathrm{cm}}^2\end{array}$Area of triangle OAP = Area of triangle OBP

$=\frac{25}{2}\times \sqrt{5}{\mathrm{cm}}^2$

Area of minor sector OACB = $=\frac{\theta }{{360}^{\circ }}\times \pi r^2$

$\begin{array}{r}=\frac{{120}^{\circ }}{{360}^{\circ }}\times 3.14(5{)}^2\\ =36.14{\mathrm{cm}}^2\end{array}$

Area of shaded region =Area of triangle OAP + Area of triangle OBP−area of minor sector OABC

$\begin{array}{r}=\frac{25\sqrt{3}}{2}+\frac{25\sqrt{3}}{2}-26.17\\ =25\sqrt{3}-26.17\\ =17.08{\mathrm{cm}}^2\end{array}$